Giải Bài 3 trang 35 SGK Toán 8 tập 1 – Chân trời sáng tạo

Đề bài

Thực hiện các phép tính sau:

a) \(\dfrac{{x + 2}}{{x – 1}} – \dfrac{{x – 3}}{{x – 1}} + \dfrac{{x – 4}}{{1 – x}}\)

b) \(\dfrac{1}{{x + 5}} – \dfrac{1}{{x – 5}} + \dfrac{{2x}}{{{x^2} – 25}}\)

c) \(x + \dfrac{{2{y^2}}}{{x + y}} – y\)

Lời giải chi tiết

a) ĐKXĐ: \(x \ne 1\)

\(\dfrac{{x + 2}}{{x – 1}} – \dfrac{{x – 3}}{{x – 1}} + \dfrac{{x – 4}}{{1 – x}}\) \( = \dfrac{{x + 2}}{{x – 1}} – \dfrac{{x – 3}}{{x – 1}} – \dfrac{{x – 4}}{{x – 1}} = \dfrac{{x + 2 – x + 3 – x + 4}}{{x – 1}} = \dfrac{{9 – x}}{{x – 1}}\)

b) ĐKXĐ: \(x \ne  \pm 5\)

\(\dfrac{1}{{x + 5}} – \dfrac{1}{{x – 5}} + \dfrac{{2x}}{{{x^2} – 25}}\) \( = \dfrac{{\left( {x – 5} \right)}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} – \dfrac{{\left( {x + 5} \right)}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} + \dfrac{{2x}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} = \dfrac{{x – 5 – x – 5 + 2x}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} = \dfrac{{2x – 10}}{{\left( {x + 5} \right)\left( {x – 5} \right)}}\)

\( = \dfrac{{2\left( {x – 5} \right)}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} = \dfrac{2}{{x + 5}}\)  

c) ĐKXĐ: \(x \ne  – y\)

\(x + \dfrac{{2{y^2}}}{{x + y}} – y\) \( = \dfrac{{x\left( {x + y} \right)}}{{x + y}} + \dfrac{{2{y^2}}}{{x + y}} – \dfrac{{y\left( {x + y} \right)}}{{x + y}} = \dfrac{{{x^2} + xy + 2{y^2} – xy – {y^2}}}{{x + y}} = \dfrac{{{x^2} + {y^2}}}{{x + y}}\)